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A181433
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Numbers k such that 11*k is 5 less than a square.
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1
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1, 4, 20, 29, 61, 76, 124, 145, 209, 236, 316, 349, 445, 484, 596, 641, 769, 820, 964, 1021, 1181, 1244, 1420, 1489, 1681, 1756, 1964, 2045, 2269, 2356, 2596, 2689, 2945, 3044, 3316, 3421, 3709, 3820, 4124, 4241, 4561, 4684, 5020, 5149, 5501, 5636, 6004
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OFFSET
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1,2
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COMMENTS
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a(k)^k+1==0 (mod 3) for k of the form 3*(2*j+1); for other forms of k, a(k)^k-1==0 (mod 3). - Bruno Berselli, Oct 29 2010
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LINKS
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FORMULA
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a(n) = (22*n*(n-1) - 5*(2*n-1)*(-1)^n + 3)/8.
a(n) = a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=1, a(2)=4, a(3)=20, a(4)=29, a(5)=61.
Sum_{i=1..n} a(i) = n*(22*n^2-15*(-1)^n-13)/24.
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MATHEMATICA
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Sort[Flatten[ Table[{((7 + 11 k)^2 - 5)/11, ((4 + 11 k)^2 - 5)/11}, {k, 0, 20, 1}]]]
Select[Range[7000], IntegerQ[Sqrt[11#+5]]&] (* Harvey P. Dale, Nov 21 2014 *)
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PROG
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(Magma) &cat[ [((4+11*k)^2-5)/11, ((7+11*k)^2-5)/11] : k in [0..23] ]; // Klaus Brockhaus, Oct 20 2010
(PARI) x='x+O('x^50); Vec(x*(1+3*x+14*x^2+3*x^3+x^4)/((1-x)^3*(1+x)^2)) \\ G. C. Greubel, Feb 25 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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Sum added and superfluous formula removed by Bruno Berselli, Oct 22 2010 - Nov 15 2010
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STATUS
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approved
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